Question

Given the following reaction:

NaCl + AgNO3 → AgCl + NaNO3
How many grams of AgCl will be produced from 7.00 g of NaCl and 95.0 g of AgNO3?
​

Answer 1

17.2

Step-by-step explanation:

Answer 2

15.76 grams of AgCl will be produced from 7.00 g of NaCl and 95.0 g of AgNO3.

Step-by-step explanation:

Balanced equation for the reaction:

NaCl + AgNO3 → AgCl + NaNO3

given:

mass of NaCl = 7 gram

mass of AgNO3 = 90 GRAM

atomic mass of NaCl = 58.44 grams/mole

atomic mass of AgNO3 = 169.87 grams/mole

number of moles =
`(mass)/(atomic mass of onemole)` 1 equation

putting the values in equation 1

number of moles of NaCl =
`(7)/(58.44)`

= 0.11 moles

number of moles of AgNO3 =
`(95)/(169.87)`

= 0.55

the limiting reagent is NaCl as it is a reactant that produces small quantity of AgCl

1 Mole of NaCl reacted to form 1 mole of AgCl

0.11 mole of NaCl will produce x moles of AgCl

`(1)/(1)` =
`(x)/(0.11)`

0.11 moles of AgCl is produced

atomic mass of AgCl = 143.32

mass = 0.11 x 143.32

= 15.76 grams of AgCl is produced.